Question

For $\alpha ϵR$ (the set of all real numbers), $a\ne -1$,
${lim}_{n\to \infty }\frac{(1^a+2^a+\dots +n^a)}{(n+1{)}^{a-1}[(na+1)+(na+2)+\dots +(na+n)]}=\frac{1}{60}$

• A. $5$
• B. $7$
• C. $\frac{-15}{2}$
• D. $\frac{-17}{2}$

Answer 1

Answer: (B), (D)

The correct options are
B $7$
D $\frac{-17}{2}$

$aϵR$
${lim}_{n\to \infty }\frac{(1^a+2^a+3^a+\dots +n^a)}{(n+1{)}^{a-1}[(na+1)+(na+2)+\dots +(na+n)]}=\frac{1}{60}$
${lim}_{n\to \infty }\frac{n^a[{\left(\frac{1}{n}\right)}^a+{\left(\frac{2}{n}\right)}^a+\dots +{\left(\frac{n}{n}\right)}^a]}{(n+1{)}^{a-1}(n^{2}a+\frac{n(n+1)}{2})}=\frac{1}{60}$
${\int }_a^{b}f\left(x\right)dx={lim}_{n\to \infty }h{\sum }_{r=0}^{n-1}f(n+rh)h=\frac{b-a}{n}$
${lim}_{n\to \infty }\frac{n^{a-1}.\frac{1}{n}{\sum }_{r=1}^n{\left(\frac{r}{n}\right)}^a}{(n+1{)}^{a-1}a+\frac{1}{2}(1+\frac{1}{n})}=\frac{1}{60}$
${lim}_{n\to \infty }{\left(\frac{1}{1+\frac{1}{n}}\right)}^{a-1}{lim}_{n\to \infty }\frac{\frac{1}{n}{\sum }_{r=1}^n{\left(\frac{r}{n}\right)}^a}{a+\frac{1}{2}(1+\frac{1}{n})}=\frac{1}{60}$
$\frac{{\int }_0^1{x}^{a}dx}{a+\frac{1}{2}(1+0)}=\frac{1}{60}\Rightarrow \frac{{\left[x^{a+1}\right]}_0^1}{(a+1)(a+\frac{1}{2})}=\frac{1}{60}\Rightarrow \frac{1}{(a+1)(2a+1)}=\frac{1}{120}$
$2a^2+3a+1=120or2a^2+3a-119=0(a=7or\frac{-17}{2})$